Question Number 153420 by alcohol last updated on 07/Sep/21 $${how}\:{many}\:{x}\:\in\mathbb{R}\:{satisfy}\:{x}^{\mathrm{99}} −\mathrm{99}{x}+\mathrm{1}=\mathrm{0} \\ $$ Answered by mr W last updated on 07/Sep/21 $${f}'\left({x}\right)=\mathrm{99}{x}^{\mathrm{98}} −\mathrm{99}=\mathrm{0} \\ $$$${x}=\pm\mathrm{1}…
Question Number 153404 by liberty last updated on 07/Sep/21 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{8}^{{x}} +\mathrm{3}^{{x}} }\:−\sqrt{\mathrm{4}^{{x}} −\mathrm{2}^{{x}} }\:=? \\ $$ Answered by EDWIN88 last updated on 07/Sep/21 $$\:\underset{{x}\rightarrow\infty}…
Question Number 153352 by liberty last updated on 06/Sep/21 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{27}^{{x}} +\mathrm{9}^{{x}} }\:−\sqrt{\mathrm{9}^{{x}} +\mathrm{3}^{{x}} }\:=? \\ $$ Answered by MJS_new last updated on 06/Sep/21 $$\underset{{x}\rightarrow+\infty}…
Question Number 22279 by A1B1C1D1 last updated on 14/Oct/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153346 by mathlove last updated on 06/Sep/21 Answered by liberty last updated on 06/Sep/21 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−\mathrm{4}}{{x}−\mathrm{2}}=\mathrm{15}\:\rightarrow\begin{cases}{{f}\left(\mathrm{2}\right)=\mathrm{4}}\\{{f}\:'\left(\mathrm{2}\right)=\mathrm{15}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}{x}^{\mathrm{2}} −{f}\left({x}\right)}{{x}−\mathrm{2}}\:=\infty \\ $$ Answered…
Question Number 153335 by liberty last updated on 06/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 87799 by M±th+et£s last updated on 06/Apr/20 $${f}\left({x}\right)=\begin{cases}{{ax}^{\mathrm{2}} +{bx}\:\:\:\:\:\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{0}}\\{{cx}^{\mathrm{2}} +{d}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}}\\{{bx}+{d}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}}\end{cases} \\ $$$${f}\left({x}\right)\:{is}\:{continuous}\:{on}\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${prove}\:{d}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}=\mathrm{2}{b} \\ $$ Commented by john santu last…
Question Number 22203 by Joel577 last updated on 13/Oct/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{x}} \:\mathrm{cos}\:{x} \\ $$ Commented by squidward last updated on 17/Oct/17 $${undefined} \\ $$ Answered…
Question Number 22165 by A1B1C1D1 last updated on 12/Oct/17 Answered by ajfour last updated on 13/Oct/17 $${let}\:\:\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:={t}\:\:\:\:\Rightarrow\:{x}^{\mathrm{2}} =\mathrm{1}−\frac{{x}}{{t}} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+\frac{{x}}{{t}}−\mathrm{2}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{{t}^{\mathrm{4}} }{\mathrm{24}}−….\right)}{\mathrm{2}{x}^{\mathrm{4}} }\right)…
Question Number 87690 by jagoll last updated on 05/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\mathrm{x}−\mathrm{sin}\:\mathrm{2x}}{\mathrm{x}−\mathrm{sin}\:\mathrm{x}} \\ $$ Commented by jagoll last updated on 05/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)−\left(\mathrm{2x}−\frac{\mathrm{8x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\mathrm{x}−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)}\:=…